Don't just guess! You want to find some function that, over a given domain, does not satisfy the definition of membership in [itex]L_{\infty}[/itex] space but does satisfy the definition of membership in [itex]L_2[/itex] space. As a hint, [itex]L_2[/itex] space comprises the set of square-integrable functions over that domain. What, in a nutshell, characterizes [itex]L_{\infty}[/itex] space?

Don't just guess! You want to find some function that, over a given domain, does not satisfy the definition of membership in [itex]L_{\infty}[/itex] space but does satisfy the definition of membership in [itex]L_2[/itex] space. As a hint, [itex]L_2[/itex] space comprises the set of square-integrable functions over that domain. What, in a nutshell, characterizes [itex]L_{\infty}[/itex] space?

[itex]L_{\infty}[/itex] = inf{C>=0: |f(x)| <=C for almost every x}
f is an element of [itex]L_{\infty}[/itex]=||f||p as p approaches infinity is finite

from here, we derived the fact that the function u that we stated is of a p-series with p<1 for L2, therefore, the integration is finite. therefore is an element of L2.

on the other hand, the function u we stated above is of p-series with p>1 for ||f||p as p approaches infinity. therefore its integration is infinite, therefore is not an element of [itex]L_{\infty}[/itex].

i just want second opinions. i just think our analysis of our function u has a lot of loopholes, given the domain constraints (unit sphere).

Forget the details. You can think of L_infinity as just 'bounded'. So you want a square integrable function on the sphere that is unbounded? Should be easy enough. I don't get your examples though. What's x? If you are working on a ball, it might be easier to use spherical coordinates.

In short, a function is in L-infinity if it is bounded throughout the domain except possibly for a set of measure zero. So, you are looking for a function that is unbounded on or within the unit sphere in R3 but the square of which is integrable over the unit sphere.

Some questions:

Why are you so bothered by the domain constraints?
This is good old R3, so you are simply looking at a volume integral over the unit sphere.

What do you mean by x in "u=((1/(x+1))^(1/c))-1 where c>2"?

If you mean [itex]x\in {\mathcal R}^3, ||x||<=1[/itex], then what does 1/(x+1) mean?

If you mean x, as in x,y,z, you should say so.

Why do you need to subtract 1?
For a finite domain, if f(x) is in Lp, then so is g(x)=f(x)+c.

Hint: That the domain is the unit sphere in R3 suggests a spherically symmetric function, thereby reducing the problem to R1.

In short, a function is in L-infinity if it is bounded throughout the domain except possibly for a set of measure zero. So, you are looking for a function that is unbounded on or within the unit sphere in R3 but the square of which is integrable over the unit sphere.

Some questions:

Why are you so bothered by the domain constraints?
This is good old R3, so you are simply looking at a volume integral over the unit sphere.

What do you mean by x in "u=((1/(x+1))^(1/c))-1 where c>2"?

If you mean [itex]x\in {\mathcal R}^3, ||x||<=1[/itex], then what does 1/(x+1) mean?

If you mean x, as in x,y,z, you should say so.

Why do you need to subtract 1?
For a finite domain, if f(x) is in Lp, then so is g(x)=f(x)+c.

Hint: That the domain is the unit sphere in R3 suggests a spherically symmetric function, thereby reducing the problem to R1.

i'm just so concerned because some integrals of functions become finite because of the domain restrictions....

ok. i'll try thinking of a spherically symmetric func, to make things simpler.